Ok, so I just finished a Discrete Math class where we learned about all types of proofs and counting and graphs, etc. So I know the basics, but never thought I'd need to use it in the real world (not anytime soon at least). However, a friend of mine showed me a puzzle game, and he said that it's always possible to win this game, even though it seems unlikely to me. I will explain the rules, which are pretty simple, and I want to know how I could prove this, or if the answer is evident using combinatorics.

Here's how the game works:
The game is a slight variation of a game called 'Poker Squares'. Anyway, you lay down (all at once) 25 random cards arranged in a 5x5 grid (no jokers). The goal of the game is to arrange all the cards so that each row is a poker hand of the following:

A straight: 5 cards arranged in order of their number (i.e. 9,10,J,Q,K, irregardless of their suite.)

A flush: any 5 cards in any order of the same suite (i.e. five hearts)

A full house: A triple and a double (i.e. three 10's and two J's.)

A straight flush: 5 cards of the same suite arranged in increasing or decreasing order (i.e. 4,5,6,7,8 all diamonds)

And thats it. The claim is that given any random 25 cards, these cards can ALWAYS be arranged so that each row is a winning poker hand (of the above mentioned hands only).

Given that there are several different types of poker hands, I think it's unlikely for there to be a particularly nice solution (that is, one that doesn't involve a lot of casework). By the way, I don't know where you got the idea that you don't use discrete math in the real world.
–
Qiaochu YuanDec 16 '10 at 21:41

No I'm sure you do use it in the real world, but I never thought I would use it, at least not any time soon..
–
maqDec 16 '10 at 21:43

1 Answer
1

Suppose you have one two of clubs, and the rest of the cards are all four and up, and all diamonds, hearts and spades (there are 33 such cards, which is more than enough). You can't use the two of clubs in any winning hand. So the claim is not true. In fact, this counterexample works even if you allow four of a kind.

How does this counterexample apply if you allow four of a kind? The row and column containing the 2C could both have a four of a kind of another rank (such as four fours and four fives).
–
SparrDec 17 '10 at 0:46

3

If you only have hearts, diamonds, and spades you can only get three of a kind.
–
Peter ShorDec 17 '10 at 2:09